ME576 Computer Control of Manufacturing Processes
Lab7 Supplement
Microprocessor (6809)
X 16 bits
Y 16 bits
U stack pointer
H stack pointer
PC
A (8 bits)
B (8 bits)
N Z V C
Index registers
(for indexed addressing modes)
Accumulator
Operations
Arithmetic

Addition
ADD
ADDB

ex) A * B  D
Subtraction
SUB
SUBB
ex)
ex)
A-MA
B-MB
Rotation and shift

Arithmetic shift to the right
ASR
ASRB

Logic or arithmetic shift to left
LSL, LSLB

A+IA
B+IB
Multiplication
MUL

ex) A + M  A
ex) B + B  B
Logic shift to the right
A-IA
B-IB
ME576 Computer Control of Manufacturing Processes
Lab7 Supplement
Logic instruction

Logic AND
AND : Logic AND between A&M or A&I
ANDB : Logic AND between B&M or B&I

Logic OR
OR : Logic OR between A&M or A&I
ORB : Logic OR between B&M or B&I

Exclusive OR
XOR : Logic XOR between A&M or A&I
XORB : Logic XOR between B&M or B&I
Incrementing, decrementing, reset

Incrementing
INC
INCB

Decrementing
DEC
DECB

Reset to zero
CLR:
CLRB:
Reset A or M to zero
Reset B to zero

Complement to 1
INV:
INVB:
One's complement A or M
Two's complement B

Complement to 2
NEG:
NEGB:
Two's complement A or M
Two's complement B
Incrementing accumulator A or M by 1
Incrementing accumulator B by 1
Data Transfer

Loading
LD
LDB

IA
II  D
IB
MA
MM  D
MB
AM
BM
D  MM
Store in memory
ST
STB
ME576 Computer Control of Manufacturing Processes
Lab7 Supplement
PLC Processor address field
RAM (8K)
Variable field
$0000 ~ $1FFFF
Inputs / outputs (8K)
Input/output, DAC, ADC, timer, etc.
$2000 ~ $3FFFF
EEPROM
EEPROM
User program
EEPROM
EPROM
PLC monitor
Processing cycle: 20 msec
Input/output Boards
Can be used in bytes and bits
read by byte directly on the input/output modules
read by bits in a working field of the PC memory
transfer if value other
than stored value
transfer of bit
Bits working field
INPUT/OUPUT board
ME576 Computer Control of Manufacturing Processes
Lab7 Supplement
RAM STRUCTURE
E.0 ~ E.FF:
256 outside inputs
A.0 ~ A.FF:
256 outside outputs
B.0 ~ B.7F:
128 internal variables/bits (B.0 ~ B.10)
T.0 ~ T.F:
32 bytes for 16 delays
C.0 ~ C.F:
32 bytes for 16 counters
B.100 ~ B.4FF:
1024 internal variables/stored bit
PC program structure
TS0
monitor
SP0-61
TS1
TS2
TS3
TS5
TS4
monitor
TF0
TF15
monitor
TSO: 20 ms scanning
TS1-TS4: 100 ms/each
TS0  TS0, TS1  TS0, TS2  TS0, TS3  TS0, TS4
ME576 Computer Control of Manufacturing Processes
Lab7 Supplement
Operators for Programming
Suffix
Operation on
E.
Input
0-FF
bit
A.
Output
0-FF
bit
S
Sequence
1-9999
EN.
Input
0-1F
byte
AN.
Output
0-1F
byte
B.
Internal variables
(Initialization)
Internal variables
0-7F
bit
80-FF
bit
B.
I.
M.
Immediate values
(decimal)
Numerical internal
variables
0-255
0-FF
Reserved variables (do not use)
B.0 to B.F : for intermediate calculations
M.0 to M.F : for special assignment
byte
ME576 Computer Control of Manufacturing Processes
Lab7 Supplement
Examples
1.
LD Loading of Accumulator A
Operator over bit.
b7
b1
b0
operand
ex.)
ex.)
LD
E.1: read and loading in accumulator A of status of input 1.
If
E.1 = 0
bit 7 = 0
If
E.1 = 1
bit 7 = 1
LD
B.10: read and loading in accumulator of A of status of
memory B.10.
B.F.
B.10
Operation over byte
ex).
LD
EN.0
:
Reading of input register 0 and storing in
accumulator register
ex).
LD
I. 128
:
Loading of immediate value into
accumulator A (decimal).
ex).
LD
I. $80
:
Loading of immediate value
accumulator A (hexadecimal).
into
ME576 Computer Control of Manufacturing Processes
2.
ST: Storage of accumulator A in memory.
ex).
ex).
ex).
3.
AND:
LD
E.1
ST
A.1
LD
E.2
E.2  (A)
ST
B.10
(A)  B.10
LD
EN.3
EN.3  (A)
ST
M.3A
(A)  M.3A
Logic and between memory and accumulator A.
E.1
ex).
E.2
B.10
( )
LD
E.1
E.1  (A)
AND
E.2
(E.1)(E.2)  (A)
ST
B.10
(A)  B.10
Lab7 Supplement
ME576 Computer Control of Manufacturing Processes
ex).
LD
EN.2
AND
EN.3
ST
AN.4
EN.2 = $79
4.
EN.3 = $85
0 1 1 1
1 0 0 1
$79
1 0 0 0
0 1 0 1
$85
0 0 0 0
0 0 0 1
$01
OR
then
Lab7 Supplement
AN.4 = $01
Logic OR between memory and Accumulator A.
E.2
A.1
( )
E.3
EN.0 = $79
EN.1 = $85
AN.0 = (?)
ME576 Computer Control of Manufacturing Processes
5.
INV
Complement of A or M.
Exercise
E.0
E.3
E.1
A.1
( )
E.2
( )
EN.0
EN.1
AN.0
EN.0 = $79
EN.1 = $85
AN.0 = (?)
Lab7 Supplement